Ask question

Ask question

All categories

2 years ago

15

Charlotte has given Ron a BIG bar of chocolate for Valentine’s day! He decides to break the bar into its individual squares to s

hare with their family. The bar is six squares long and thirteen squares high. How many times will he have to break the bar?

1 answer:

Cerrena [4.2K]2 years ago

6 0

Length of bar= 6 units
Width of bar= 13 units
Formula for area of rectangular chocolate bar:
A= L x W
Put values
A= 6 x 13
A=78 square units
So he will break the bar into 78-1=77 times

Answer: 77 times

You might be interested in

How can you express the properties of operations to evaluate this expression 18 + 4(28)

Akimi4 [234]

Answer:

I don't know if I'm understanding your question correctly. But I'll try...

In this problem you need to remember PEMDAS!

Parentheses, Exponents, Multiplication, Division, Addition, and Subtraction.

18 + 4(28) -First you would multiply 4 by 28 since the 28 is in parentheses.

18 + 112 -Now you would add to find your answer

= 130

I hope this helps!

-Mikayla

3 0

2 years ago

Given the function g(x) = –x^2+ 4x + 9, determine the average rate of change of

Ghella [55]

Answer:

9

Step-by-step explanation:

4 0

2 years ago

Garth estimated the height of the door to his classroom in meters.what is a reasonable estimate?

insens350 [35]

A reasonable estimate is an educated guess made by an observer based on that person's current knowledge and following observations. For example, If Garth knew that he was 1.2 meters tall, and he observed that the door was approximately twice his own height, Garth could then make the reasonable estimate that the door is about 2.4 meters tall.

8 0

3 years ago

Match each circular flower bed on the left to its circumference and its area on the right. Some answer options on the right will

marin [14]

Answer:

Bed A --- Third Option

Bed B --- Fifth Option

Step-by-step explanation:

Flower Bed A)

The formula for circumference is:

$C=2\pi r$

The diameter of A is 10 feet. So, its radius is 5 feet.

Substitute 5 for r. So, the circumference of Bed A is:

$C=2\pi (5)=10\pi\text{ feet}$

The formula for the area of a circle is:

$A=\pi r^2$

Substitute 5 for r:

$A=\pi (5)^2$

Square and simplify:

$A=25\pi \text{ square feet}$

So, the circumference of Bed A is 10π feet and its area is 25π square feet.

Therefore, match Bed A to the third option.

Flower Bed B)

The radius of Bed B is 6 feet. Again, we can use the circumference and area formulas. Substitute 6 for r for circumference:

$C=2\pi (6)$

Multiply:

$C=12\pi\text{ feet}$

Substitute 6 for r for the area formula:

$A=\pi(6)^2$

Square and simplify:

$A=36\pi\text{ square feet}$

So, Bed B has a 12π feet circumference and a 36π square feet area.

Therefore, match Bed B to the fifth option.

And we're done!

8 0

3 years ago

A large corporation starts at time t = 0 to invest part of its receipts continuously at a rate of P dollars per year in a fund f

Andrews [41]

Answer:

$A = \frac{P}{r}\left( e^{rt} -1 \right)$

Step-by-step explanation:

This is <em>a separable differential equation</em>. Rearranging terms in the equation gives

$\frac{dA}{rA+P} = dt$

Integration on both sides gives

$\int \frac{dA}{rA+P} = \int dt$

where $c$ is a constant of integration.

The steps for solving the integral on the right hand side are presented below.

$\int \frac{dA}{rA+P} = \begin{vmatrix} rA+P = m \implies rdA = dm\end{vmatrix} \\\\\phantom{\int \frac{dA}{rA+P} } = \int \frac{1}{m} \frac{1}{r} \, dm \\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \int \frac{1}{m} \, dm\\\\\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |m| + c \\\\&\phantom{\int \frac{dA}{rA+P} } = \frac{1}{r} \ln |rA+P| +c$

Therefore,

$\frac{1}{r} \ln |rA+P| = t+c$

Multiply both sides by $r.$

$\ln |rA+P| = rt+c_1, \quad c_1 := rc$

By taking exponents, we obtain

$e^{\ln |rA+P|} = e^{rt+c_1} \implies |rA+P| = e^{rt} \cdot e^{c_1} rA+P = Ce^{rt}, \quad C:= \pm e^{c_1}$

Isolate $A$ .

$rA+P = Ce^{rt} \implies rA = Ce^{rt} - P \implies A = \frac{C}{r}e^{rt} - \frac{P}{r}$

Since $A = 0$ when $t=0$ , we obtain an initial condition $A(0) = 0$ .

We can use it to find the numeric value of the constant $c$ .

Substituting $0$ for $A$ and $t$ in the equation gives

$0 = \frac{C}{r}e^{0} - \frac{P}{r} \implies \frac{P}{r} = \frac{C}{r} \implies C=P$

Therefore, the solution of the given differential equation is

$A = \frac{P}{r}e^{rt} - \frac{P}{r} = \frac{P}{r}\left( e^{rt} -1 \right)$

4 0

2 years ago